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96.5 amperes current is passed through the molten $\mathrm{AlCl}_3$ for 100 seconds. The mass of aluminium deposited at the cathode is (atomic weight of $\mathrm{Al}=27 \mathrm{u}$ )
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The correct answer is:
$0.90 \mathrm{~g}$
Given, $I=96.5 \mathrm{~A}$
$t=100 \mathrm{~s}$
$\Rightarrow \quad m=?$
Atomic weight of $\mathrm{Al}=27 \mathrm{u}$
$Q=I \times t=96.5 \times 100$
$=9650 \mathrm{C}$
$\begin{aligned} & \mathrm{AlCl}_3 \longrightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ & \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al} \\ & 96500 \times 3 \mathrm{C} \text { of electricity deposits, } 1 \text { mole } \mathrm{Al} .\end{aligned}$
$\therefore 9650 \mathrm{C}$ will deposit $\mathrm{Al}=\frac{9650}{96500 \times 3}=0.0333 \mathrm{~mol}$.
Weight of $\mathrm{Al}$ deposited $=n \times$ molar mass
$\begin{aligned} & =0.0333 \times 27 \\ & \cong 0.90 \mathrm{~g}\end{aligned}$
$t=100 \mathrm{~s}$
$\Rightarrow \quad m=?$
Atomic weight of $\mathrm{Al}=27 \mathrm{u}$
$Q=I \times t=96.5 \times 100$
$=9650 \mathrm{C}$
$\begin{aligned} & \mathrm{AlCl}_3 \longrightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ & \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al} \\ & 96500 \times 3 \mathrm{C} \text { of electricity deposits, } 1 \text { mole } \mathrm{Al} .\end{aligned}$
$\therefore 9650 \mathrm{C}$ will deposit $\mathrm{Al}=\frac{9650}{96500 \times 3}=0.0333 \mathrm{~mol}$.
Weight of $\mathrm{Al}$ deposited $=n \times$ molar mass
$\begin{aligned} & =0.0333 \times 27 \\ & \cong 0.90 \mathrm{~g}\end{aligned}$
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